Question #268057

Block 1 of mass π‘š1 = 2 π‘˜g is placed on block 2 of mass π‘š2 = 3 π‘˜g which is then placed on a table. A string connecting block 2 to a hanging mass 𝑀 passes over a pulley attached to one end of the table, as shown in Fig. 1. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and block 2 are πœ‡1s = 0.2, πœ‡1k = 0.1 and between block 2 and the tabletop are πœ‡2s = 0.3, πœ‡2k = 0.15


(a) (Marks: 3) Draw the complete free body diagram of the system and determine the largest value of 𝑀 for which the blocks can remain at rest.

(b) (Marks: 4) Now suppose that 𝑀 = 2.5 π‘˜g which is large enough so that the hanging block descends and block 1 slips on block 2. Draw the complete free body diagram of the system and determine each of the followings.

i. The magnitude π‘Ž1of the acceleration of block 1.

ii. The magnitude π‘Ž2 of the acceleration of block 2.




Expert's answer


a)



The maximum friction force on the blocks on the table:

Ffr2max=ΞΌs2N2=ΞΌs2(m1+m2)gF_{fr2max}=\mu_{s2}N_2=\mu_{s2}(m_1+m_2)g

which is balanced by the weight of the hanging mass:

Mg=ΞΌs2(m1+m2)gMg=\mu_{s2}(m_1+m_2)g

then:

M=ΞΌs2(m1+m2)=0.3g(2+3)=14.7M=\mu_{s2}(m_1+m_2)=0.3g(2+3)=14.7 kg


b)



i)

Ffr1=ΞΌk1m1g=m1a1F_{fr1}=\mu_{k1}m_1g=m_1a_1

a1=ΞΌk1g=0.1g=0.98a_1=\mu_{k1}g=0.1g=0.98 m/s2


ii)

for the two blocks:

Mgβˆ’T=Ma2Mg-T=Ma_2

Tβˆ’Ffr1βˆ’Ffr2=m2a2T-F_{fr1}-F_{fr2}=m_2a_2


a2=Mβˆ’ΞΌk1m1βˆ’ΞΌk2(m1+m2)M+m2g=2.5βˆ’0.1β‹…2βˆ’0.15(2+3)2.5+3g=2.76a_2=\frac{M-\mu_{k1}m_1-\mu_{k2}(m_1+m_2)}{M+m_2}g=\frac{2.5-0.1\cdot2-0.15(2+3)}{2.5+3}g=2.76 m/s2

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